pt bậc 1 đối với sinx cosx
tanx+2cotx-3=0
tìm TXĐ của hàm số
y=5tan^2x+4cot^2x
y=2tanx+3/sin2x
y=x+1/sin2x+3
pt bậc 1 đối với sinx cosx
tanx+2cotx-3=0
tìm TXĐ của hàm số
y=5tan^2x+4cot^2x
y=2tanx+3/sin2x
y=x+1/sin2x+3
\[\begin{array}{l}
+ )\,\,\,\tan x + 2\cot x – 3 = 0\\
DK:\,\,\,\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\
\Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\
\Rightarrow pt \Leftrightarrow \tan x + 2.\frac{1}{{\tan x}} – 3 = 0\\
\Leftrightarrow {\tan ^2}x – 3\tan x + 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = 2
\end{array} \right. \Leftrightarrow ….\\
Tim\,\,\,TXD\,\,\,cua\,\,\,ham\,\,so:\\
+ )\,\,\,y = 5{\tan ^2}x + 4{\cot ^2}x\\
DK:\,\,\,\left\{ \begin{array}{l}
\cos x \ne 0\\
\sin x \ne 0
\end{array} \right. \Leftrightarrow \sin 2x \ne 0 \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\
\Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\
+ )\,\,y = \frac{{2\tan x + 3}}{{\sin 2x}}\\
DK:\,\,\,\left\{ \begin{array}{l}
\cos x \ne 0\\
\sin 2x \ne 0
\end{array} \right. \Leftrightarrow \Leftrightarrow 2x \ne k\pi \Leftrightarrow x \ne \frac{{k\pi }}{2}\\
\Rightarrow D = R\backslash \left\{ {\frac{{k\pi }}{2}} \right\}.\\
+ )\,\,\,y = \frac{{x + 1}}{{\sin 2x + 3}}\\
\sin 2x + 3 > 0\,\,\forall x\\
\Rightarrow D = R.
\end{array}\]